Rotation of rigid bodies: yo-yo

AI Thread Summary
The discussion focuses on calculating the final velocity of a yo-yo after it falls a height h, using Newton's second law. The user applies the equation F = ma to the combined mass of the yo-yo and derives the net force and torque equations. They calculate the moment of inertia for the two cylinders and derive a formula for linear acceleration, which turns out to be constant. The conversation highlights that since acceleration is constant, the final velocity can be determined using the relationship between distance fallen and constant acceleration. The final insight connects gravitational acceleration to the derived linear acceleration, suggesting a straightforward approach to find the final velocity.
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consider a yo-yo consisting of two cylinders of radius R1, (combined) mass M1, glued to another smaller cylinder of radius R2, mass M2. find the final velocity after falling a height h using Newton's second law. assume the string is vertical.

Chapter1229.gif


the inner radius R2 is R0, the outer radius R1 is R in this picture

so, i started off by applying f = ma to M2..

Fnet, y = (M1 + M2) - T = (M1 + M2)(a)cm

tnet, ext = T(R1) = I(alpha), alpha = (a)cm/R1

i calculated the moment of inertial to be the sum of the moments of inertia of the cylinders..

I = 1/2*(M1*R1^2 + M2*R2^2)

solving for (a)cm, i get a very nasty formula..

a = (M1 + M2)g / ( ( I/(2R1^2)) + (M1 + M2))

my original plan was to integrate this wrt time to get a velocity.
but this formula doesn't involve anything that changes with time.. so how do i obtain the final velocity? am i even on the right track?
 
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It looks ok.
As you noticed, the expression for a doesn't involve anything that changes wrt time. In other words, a is constant!

If you had an object that was falling under gravity, what would its velocity be after having fallen a distance h?

So then, what if g = a?
 
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